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Chapter 7: Problem 46

Evaluate the following integrals. $$\int \frac{x^{3}}{\left(x^{2}-16\right)^{3 / 2}} d x, x<-4$$

Short Answer

Expert verified
Question: Evaluate the integral, given the interval x < -4: $$\int \frac{x^{3}}{\left(x^{2}-16\right)^{3 / 2}} d x$$ Answer: $$-\frac{x^2}{\sqrt{x^2 - 16}} + \sqrt{x^2 - 16}$$

Step by step solution

01

Substitution

Let's make a substitution, \(u = x^2 - 16\). Then, $$\frac{du}{dx} = 2x \Rightarrow dx = \frac{du}{2x}.$$ Now, the integral becomes: $$\int \frac{x^3}{(u)^{3 / 2}} \frac{du}{2x}$$ Step 2: Simplify the integral
02

Simplify

Now, simplify the integral by canceling the \(x\) term and the \(2\) in the denominator: $$\frac{1}{2} \int \frac{x^2}{u^{3/2}} du$$ Substitute back \(u = x^2 - 16\): $$\frac{1}{2} \int \frac{x^2}{(x^2 - 16)^{3/2}} du$$ Now, with the same substitution as before, change the variable back to \(x\) and rewrite the integral in terms of \(x\): $$\frac{1}{2} \int \frac{x^2}{(x^2 - 16)^{3/2}} dx$$ Step 3: Evaluating the integral using integration by parts
03

Integration by Parts

Before integration by parts, rewrite the integral as: $$ \frac{1}{2}\int x^2(x^2-16)^{-3/2}dx. $$ Now, let \(u = x^2\) and \(dv = (x^2 - 16)^{-3/2} dx\). Then, $$du = 2x dx, \quad v = -\frac{2}{\sqrt{x^2 - 16}}$$ Now, use integration by parts formula: \(\int u dv = uv - \int v du\): $$\frac{1}{2}\int x^2(x^2-16)^{-3/2}dx = \frac{1}{2}\left(-x^2\frac{2}{\sqrt{x^2 - 16}} - \int \left(-\frac{2}{\sqrt{x^2 - 16}}\right) 2x dx\right)$$ Step 4: Simplify and Find the Antiderivative
04

Antiderivative

$$\frac{1}{2}\int x^2(x^2-16)^{-3/2}dx =\frac{1}{2}\left(-\frac{2x^2}{\sqrt{x^2 - 16}} + 4\int \frac{x}{\sqrt{x^2 - 16}}dx\right)$$ $$= -\frac{x^2}{\sqrt{x^2 - 16}} + 2\int \frac{x}{\sqrt{x^2 - 16}}dx$$ Now, evaluating the remaining integral using substitution \(v = x^2 - 16\), we have \(x = \frac{dv}{2}\) and \(dx = \frac{dv}{2}\) $$= -\frac{x^2}{\sqrt{x^2 - 16}} + 2\int \frac{dv}{2\sqrt{v}} = -\frac{x^2}{\sqrt{x^2 - 16}} + \sqrt{v} + C$$ $$= -\frac{x^2}{\sqrt{x^2 - 16}} + \sqrt{x^2 - 16} + C$$ Step 5: Evaluate the Limit for x < -4
05

Fundamental Theorem of Calculus and Limit

Since we're evaluating the integral for \(x<-4\), the singularities do not interfere with the interval of interest. Thus the antiderivative turns out to be $$F(x) = -\frac{x^2}{\sqrt{x^2 - 16}} + \sqrt{x^2 - 16} + C.$$ So, the final solution for the integral is: $$\int \frac{x^{3}}{\left(x^{2}-16\right)^{3 / 2}} d x = -\frac{x^2}{\sqrt{x^2 - 16}} + \sqrt{x^2 - 16}.$$

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